Question

Which equation could generate the curve in the graph below? y = –2x2 + 3x – 5 y = –2x2 – 4x – 2 y = –2x2 – 16x– 28 y = –2x2 +16x –28

Answer 1

In order to find which equation could generate the curve, we can take each option and verify if delta if greater than zero because we have 2 intersection points with OX and if those intersection points are both negative (the intersection point are the solution of the equation).

First option: delta = 3^2-4*(-2)*(-5) = 9-40<0 not a good option
Second option: delta = 4^2-4*(-2)*(-2) = 16-16=0<0 not a good option
Third option: delta = 16^2-4*(-2)*(-28) = 256-224 = 32
x1,2 = (16+-√32)/(-4) = -4-+√2
Both values are negative and delta<0 so this is a good solution.

Fourth option: delta = 16^2-4*(-2)*(-28) = 256-224 = 32
x1,2 = (-16+-√32)/(-4) = 4-+√2. Just one solution is negative the other one is positive. Not a good solution.

The final equation is:
y = –2x^2 – 16x– 28

Answer 2

The equation could generate the curve in the graph is $\rm y = -2x^2 - 16x-28$.

We have to determine

Which equation could generate the curve in the graph below?

What is the general equation of the parabola in quadratic form?

The general equation of the parabola in quadratic form is;

$\rm y = ax^2+bx+c$

Where the vertex of the parabola is (h, k).

• In the given graph the parabola opens downwards and the vertex is in the second quadrant.

• The vertex's x coordinate (h) is negative, while the they-coordinate (k) is positive.

• The equation which generates the given graph is x-coordinate (h) of the vertex is negative, while y-coordinate (k) is positive.

Therefore;

The equation is;

$\rm h=\dfrac{-16}{-2\times -2}\\\\h =\dfrac{-16}{4}\\\\h = -4$

Substitute -4 for x in the function;

$\rm k =-2x2 +16x -28\\\\k = -2(-4)^2-16(-4)-28\\\\k = -32+64-28\\\\k =4$

Hence, the equation could generate the curve in the graph is $\rm y = -2x^2 - 16x-28$.

To know more about Parabola click the link given below.

https://brainly.com/question/4059031